Explanation:
The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.
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Answer:</h3>
0.10 L
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Explanation:</h3>
The concentration of glucose is given as 180 g/L
The mass of glucose is 18 g
- Concentration in g/L is calculated by dividing mass of the solute by the volume of the solution.
- When calculating molarity on the other hand, we divide number of moles of the solute by the volume of the solution.
- Concentration in g/L = Mass of solute ÷ Volume
Rearranging the formula,
Volume = Mass of the solute ÷ concentration
= 18 g ÷ 180 g/L
= 0.10 L
Therefore, volume of water is 0.10 L
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Answer:

Explanation:
Mass ( m ) = 52 kg
Acceleration ( a ) = 5 m / s²
Force ( F ) = ?
According to Newton's second law of motion ,



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Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.